A spherical balloon is being inflated so that its – Class 12 Mathematics | ConceptFirst

Example · Example 2

900\ \text{cm}^3/\text{s}

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✓ Free question

Let $r$ be the balloon's radius at time $t$ and $V=\tfrac{4}{3}\pi r^3$ its volume, with $\dfrac{dV}{dt}=900\ \text{cm}^3/\text{s}$ given.

Differentiating with respect to $t$ :

\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}.

At $r = 15\ \text{cm}$ : $\quad 900 = 4\pi(15)^2\dfrac{dr}{dt} = 900\pi\,\dfrac{dr}{dt}$ , so

\frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi}\ \text{cm/s}.

✓Final answer

\dfrac{dr}{dt} = \dfrac{1}{\pi}\ \text{cm/s}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$10\ \text{cm}^3/\text{sec}$
(a) $1{\cdot}5\ \text{cm}^2/\text{sec}$
(b) $1{\cdot}8\ \text{cm}^2/\text{sec}$
(c) $2\ \text{cm}^2/\text{sec}$
(d) $1{\cdot}25\ \text{cm}^2/\text{sec}$
›Reveal solutionSolution

Relate $\tfrac{dS}{dt}$ to $\tfrac{dV}{dt}$ through $\tfrac{dr}{dt}$ ; the surface area changes at $\tfrac{20}{r}=1{\cdot}25\ \text{cm}^2/\text{sec}$ .

This is a classic related-rates problem from the NCERT/CBSE Class 12 application of derivatives chapter.

Volume $V=\tfrac43\pi r^3$ , so

$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 10 \Rightarrow \frac{dr}{dt} = \frac{10}{4\pi r^2}.$

Surface area $S = 4\pi r^2$ , so

$\frac{dS}{dt} = 8\pi r\frac{dr}{dt} = 8\pi r\cdot\frac{10}{4\pi r^2} = \frac{20}{r}.$

At $r=16$ cm:

$\frac{dS}{dt} = \frac{20}{16} = 1{\cdot}25\ \text{cm}^2/\text{sec}.$

✓Final answer
$= 1{\cdot}25\ \text{cm}^2/\text{sec}$
CBSE 2019Set ANNUAL1 markMCQ
$Q. The rate of increase of a side of a square is 1 cm/sec. The rate of increase of area of the square, when length of a side of the square is 2 cm, is$
(a) 4 cm^2/sec
(b) 8 cm^2/sec
(c) 1 cm^2/sec
(d) 2 cm^2/sec
›Reveal solutionSolution

Relate $dA/dt$ to $ds/dt$ via $A=s^2$ , then substitute $s=2$ .

Let $s$ be the side length, $A=s^2$ the area. Given $\dfrac{ds}{dt}=1$ cm/sec.

Differentiate $A=s^2$ w.r.t. time:

$\dfrac{dA}{dt}=2s\dfrac{ds}{dt}$

At $s=2$ cm:

$\dfrac{dA}{dt}=2(2)(1)=4 \text{ cm}^2/\text{sec}$

✓Final answer
$4\text{ cm}^2/\text{sec}$

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